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Re: st: psmatch2 pstest output interpretation


From   "Austin Nichols" <[email protected]>
To   [email protected]
Subject   Re: st: psmatch2 pstest output interpretation
Date   Fri, 11 Jan 2008 15:57:47 -0500

Claire--
I guess I still don't understand what you are trying to do.  The
-psmatch2- command gives you the t-stat for the mean difference in
outcomes for the treatment group when treated and (counterfactually)
untreated (where the null is no diff).  The -pstest- command is
intended for a different test, for checking covariate imbalance--to
see if your matching made the treatment and control groups more alike,
looking at vars one by one, rather than all together like a
-hotelling- test or a regression of treatment status on covariates
(but for the "fake" matched data now) would. So I don't know what your
use of -pstest- is intended to do.

webuse nlswork, clear
xi i.race i.ind i.occ
loc x "union coll age ten not_s c_city south nev_m _I*"
g u=uniform()
sort u
psmatch2 `x', out(ln_w) ate
* ATT comparison a bit like this next one... with bigger SE
ttest ln=_ln if _treat==1 & _supp==1
* don't know what this kind of command would be for...
pstest _ln
* but it's more like this kind of thing:
reg _ln _treat [iw=_wei] if _supp==1
* next bit is what pstest is for: seeing if T&C more alike on X's
pstest age ten

You can look up the formulas in the ado code, or in the refs...  but why?

On Jan 11, 2008 2:04 PM, Claire Kamp Dush <[email protected]> wrote:
> Hi Austin and statalist, I am only interested in the mean comparison
> across matched observations for the treatment group.  So, in the example
> below, the treatment group (i.e. marcohdis23 = 1) is the group who
> dissolves a cohabiting union, and the control group, or non-treated
> group, (i.e. marcohdis23 == 0) is the divorced group.  So, my question
> is why are there two different t test statistics calculated by psmatch2
> and pstest for the matched sample?  Why do I get the same t statistic
> for the unmatched sample but different t test?  Thanks Austin for
> replying, Claire
>
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